72 Trisection of an Angle. [Fes. 
Let ABC (fig. 2) be any given rectilineal angle greater than half a right 
angle. Having placed it for the vertical angle of the triangle ABG, 
it is required to draw the base AG, so that of the interior angles it 
shall make with BA and BC, at the base of the triangle ABG, the one 
shall be double of the other. 
From the centre B at any distance BA describe the circle ADE ; 
again, from the centre B at twice the distance BA describe the arch of a 
circle FH, cutting BC, in F; also from the centre A at three times the 
distance BA, mark the point C in the line BC ; divide the segment FC 
into three equal parts (9.6); make FG equal to one-third part of FC 
(3.1); through Gdraw GH at right angles to BC (11.1), meeting the 
arch FH in the point H; jom BH and GA; the line GA is so drawn 
that BAG, one of the angles at the base of the triangle ABG, is double 
of BGA, the other angle at the base. 
Because the two straight lines BH and AG and the circle ADE in- 
tersect in D, the two sides BD, BA of the triangle ABD, being radii 
of the circle ADE, are equal to one another (11.def.) ; also BH, which 
is equal to BF (11.def.) and double of BA or BD, is bisected in D; 
again, because BGH is a right angle subtended by BH, it is an angle 
in half the circle, having BH for its diameter and DB for its radius 
(31.8); and because GD joins the vertex of the right angle BGH and 
D, the point of bisection ofthe diameter, it is equal to DB (11.def.): 
Now because DG is equal to DB, the angle DBG is equal to the angle 
DGB (5.1); and they are the two interior and opposite angles to BDA, 
the exterior angle of the triangle BDG, therefore BDA is equal to 
them both (32.1), and double of either of them, that is, it is double of 
DGB; but the angle BAD is equalto the angle BDA, because BD is 
equal to BA (5.1), therefore BAD, that is BAG, is double of DGB, that 
is, AGB. Wherefore the base AG is drawn so that the angle BAG, one 
of the angles at the base of the triangle ABG, is double of BGA, the 
other angle at the base. Which was required to be done. 
Note.—The truth of the above demonstration rests upon the straight 
lines BH and AG and the circle ADE having one common intersection; 
but as the circle and any two right lines have not of necessity one 
common intersection, it may perhaps be objected to, on the ground, 
that though it states the fact, it does not prove the intersection of the 
circle ADE and the right lines BH and AG in the common point D. 
To remove that objection, the following demonstration is given. 
The construction being the same as above, instead of joing GA, 
proceed thus :—join BH, and let BH cut the circle ADE in the point D; 
join GD and DA; AD, DG are in the same straight line, and AG, the 
